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Show that the two straight lines $x^2(\tan^2 (\theta)+\cos^2 (\theta))-2xy\tan (\theta)+y^2.\sin^2 (\theta)=0$ make with x axis such that the difference of their tangents is $2$.

My Attempt: $$x^2(\tan^2 (\theta) +\cos^2 (\theta))-2xy\tan (\theta) + y^2 \sin^2 (\theta)=0$$

Let $y-m_1x=0$ and $y-m_2x=0$ be the two lines represented by the above equation. Their combined equation is: $$(y-m_1x)(y-m_2x)=0$$ $$y^2-(m_1+m_2)xy+(m_1m_2)x^2=0$$

How do I proceed further?

pi-π
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  • Not saying your approach is not right, but note that their "combined equation" can also be $\min(|y-m_1x|,|y-m_2x|)=0$, or other expressions. I would go the other way around, i.e. replace $y=ax$ in the equation and use some trig to simplify, and reduce to an equation on $a$. – Tal-Botvinnik Apr 23 '18 at 11:57
  • Also, what makes you that the lines pass through (0,0)? – Tal-Botvinnik Apr 23 '18 at 11:58
  • @ Tal-Botvinnik, Clearly the given equation is in the form of $ax^2+2hxy+by^2=0$ which is the homogeneous equation of second degree. And obviously the two lines represented by it pass through the origin. – pi-π Apr 23 '18 at 12:03

1 Answers1

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We have $$\left(\dfrac yx\right)^2\sin^2\theta-\dfrac yx(2\tan\theta)+\tan^2\theta+\cos^2\theta=0$$

If $m_1,m_2$ are the two roots

$m_1m_2=\dfrac{\tan^2\theta+\cos^2\theta}{\sin^2\theta}=\dfrac{s^2+c^4}{c^2s^2}$

$m_1+m_2=\dfrac{2s}{cs^2}=\dfrac2{cs}$

$$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2=4\cdot\dfrac{1-(s^2+c^4)}{c^2s^2}=\dfrac{4c^2(1-c^2)}{c^2s^2}=4$$

lab bhattacharjee
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